If `int(dx)/((x+2)(x^(2)+1)) = alog|1+x^(2)|+btan^(-1)x+ 1/5log|x+2|+C`, then

Options

A. `a=(-1)/(10), b=(-2)/(5)`
B. `a=(1)/(10), b=(-2)/(5)`
C. `a=(-1)/(10),b=(2)/(5)`
D. `a=(1)/(10), b=(2)/(5)`

Answer

C

Explanation

Let `I=int(dx)/((x+1)(x^(2)+1))`
By using partial fraction method, write
`(1)/((x+2)(x^(2)+1))=(A)/(x+2)+(Bx+C)/(x^(2)+1)`
`=(Ax^(2)+A+Bx^(2)+Cx+2Bx+2C)/((x+2)(x^(2)+1))`
`1=x^(2)(A+B)+x(C+2B)+(A+2C)`
On comparing the coefficients of `x^(2),x` and the constant on both sides, we get
`A+B=0`
`2B+C=0`
`A+2C=1`
`rArr" "A=-B`
`{:(-2A+C=0),(2A+4C=2),(_),(" "5C=2):}`
`rArr" "C=(2)/(5)`
`therefore" "A=1-2C=1-2xx(2)/(5)=1-(4)/(5)=(1)/(5) rArr B=-A=-(1)/(5)`
`therefore " "l=(1)/(5)int(1)/(x+2)dx+(-1)/(5)int(x)/(x^(2)+1)dx+(2)/(5)int(dx)/(x^(2)+1)`
`=(1)/(5)log|x+2|+((-1)/(10))log|x^(2)+1|+(2)/(5)tan^(-1)x+C`
On comparing with
`a log |1+x^(2)|+b tan^(-1)x+(1)/(5)log|x+2|+C,` we get
`a=(-1)/(10) and b=(2)/(5)`